∫01x(x2−2)2dx\int\limits_{0}^{1} x(x^2-2)^2 dx0∫1x(x2−2)2dx
(x2−2)2=x4−2x2−2x2+4(x^2-2)^2=x^4-2x^2-2x^2+4(x2−2)2=x4−2x2−2x2+4
=x4−4x2+4x^4-4x^2+4x4−4x2+4
x(x2−2)2=x(x4−4x2+4)x(x^2-2)^2=x(x^4-4x^2+4)x(x2−2)2=x(x4−4x2+4)
=x5−4x3+4xx^5-4x^3+4xx5−4x3+4x
∫01x(x2−2)2dx=∫01x5−4x3+4xdx\int\limits_{0}^{1} x(x^2-2)^2 dx = \int\limits_{0}^{1} x^5 - 4x^3 + 4x dx0∫1x(x2−2)2dx=0∫1x5−4x3+4xdx
=( x66−x4+2x2)01\left( \;\frac{x^6}{6} - x^4+2x^2 \right)_0^1(6x6−x4+2x2)01
= ( (1)66−(1)4+2(1)2)−( (0)66−(0)4+2(0)2)\left( \;\frac{(1)^6}{6} - (1)^4 +2(1)^2 \right) - \left( \;\frac{(0)^6}{6} - (0)^4+2(0)^2 \right)(6(1)6−(1)4+2(1)2)−(6(0)6−(0)4+2(0)2)
= 76−0= 76\;\frac{7}{6} - 0 = \;\frac{7}{6}67−0=67
∴1 16\therefore 1\;\frac{1}{6}∴161